Question

A unit vector perpendicular to the plane passing through the points whose position vectors are 2i –­ j + 5k, 4i + 2j + 2k and 2i +­ 4j + 4k is-

• A. $\frac { 6 i + j + 5 k } { \sqrt { 62 } }$
• B.
• C. $\frac { - 6 \mathrm { i } + \mathrm { j } + 5 \mathrm { k } } { \sqrt { 62 } }$
• D. $\frac { 9 i + j - 5 k } { \sqrt { 107 } }$

Answer 1

Answer: (A)

$\frac { 6 i + j + 5 k } { \sqrt { 62 } }$

Answer 2

If A, B, C are the three points

$\overline { \mathrm { AB } }$ = 2i + 3j – 3k, $\overline { \mathrm { AC } }$= 5j – k

Required unit vector = $\frac { \overline { \mathrm { AB } } \times \overline { \mathrm { AC } } } { | \overline { \mathrm { AB } } \times \overline { \mathrm { AC } } | }$ = $\frac { 6 \mathrm { i } + \mathrm { j } + 5 \mathrm { k } } { \sqrt { 62 } }$